MACLAURIN’S SERIES 71
A
- Use Maclaurin’s series to determine the
expansion of (3+ 2 t)^4.
[
81 + 216 t+ 216 t^2 + 96 t^3 + 16 t^4
]
8.5 Numerical integration using
Maclaurin’s series
The value of many integrals cannot be determined
using the various analytical methods. In Chapter
45, the trapezoidal, mid-ordinate and Simpson’s
rules are used to numerically evaluate such inte-
grals. Another method of finding the approximate
value of a definite integral is to express the func-
tion as a power series using Maclaurin’s series, and
then integrating each algebraic term in turn. This is
demonstrated in the following worked problems.
Problem 11. Evaluate
∫ 0. 4
0. 1 2e
sinθdθ, correct
to 3 significant figures.
A power series for esinθ is firstly obtained using
Maclaurin’s series.
f(θ)=esinθ f(0)=esin 0=e^0 = 1
f′(θ)=cosθesinθ f′(0)=cos 0 esin 0=(1)e^0 = 1
f′′(θ)=(cosθ)(cosθesinθ)+(esinθ)(−sinθ),
by the product rule,
=esinθ(cos^2 θ−sinθ);
f′′(0)=e^0 (cos^20 −sin 0)= 1
f′′′(θ)=(esinθ)[(2 cosθ(−sinθ)−cosθ)]
+(cos^2 θ−sinθ)(cosθesinθ)
=esinθcosθ[−2 sinθ− 1 +cos^2 θ−sinθ]
f′′′(0)=e^0 cos 0[(0− 1 + 1 −0)]= 0
Hence from equation (5):
esinθ=f(0)+θf′(0)+
θ^2
2!
f′′(0)+
θ^3
3!
f′′′(0)+···
= 1 +θ+
θ^2
2
+ 0
Thus
∫ 0. 4
0. 1
2esinθdθ=
∫ 0. 4
0. 1
2
(
1 +θ+
θ^2
2
)
dθ
=
∫ 0. 4
0. 1
(2+ 2 θ+θ^2 )dθ
=
[
2 θ+
2 θ^2
2
+
θ^3
3
] 0. 4
0. 1
=
(
0. 8 +(0.4)^2 +
(0.4)^3
3
)
−
(
0. 2 +(0.1)^2 +
(0.1)^3
3
)
= 0. 98133 − 0. 21033
= 0. 771 , correct to 3 significant figures
Problem 12. Evaluate
∫ 1
0
sinθ
θ
dθ using
Maclaurin’s series, correct to 3 significant
figures.
Let f(θ)=sinθ f(0)= 0
f′(θ)=cosθ f′(0)= 1
f′′(θ)=−sinθ f′′(0)= 0
f′′′(θ)=−cosθ f′′′(0)=− 1
fiv(θ)=sinθ fiv(0)= 0
fv(θ)=cosθ fv(0)= 1
Hence from equation (5):
sinθ=f(0)+θf′(0)+
θ^2
2!
f′′(0)+
θ^3
3!
f′′′(0)
+
θ^4
4!
fiv(0)+
θ^5
5!
fv(0)+···
= 0 +θ(1)+
θ^2
2!
(0)+
θ^3
3!
(−1)
+
θ^4
4!
(0)+
θ^5
5!
(1)+···
i.e. sinθ=θ−
θ^3
3!
+
θ^5
5!
−···
Hence
∫ 1
0
sinθ
θ
dθ=
∫ 1
0
(
θ−
θ^3
3!
+
θ^5
5!
−
θ^7
7!
+···
)
θ
dθ
=
∫ 1
0
(
1 −
θ^2
6
+
θ^4
120
−
θ^6
5040
+···
)
dθ